TEST BANK FOR A First Course in Differential Equations with Modeling Applications 9th

Complete Solutions Manual
A First Course in Differential Equations
with Modeling Applications
Ninth Edition
Dennis G. Zill
Loyola Marymount University
Differential Equations with
Boundary-Vary Problems
Seventh Edition
Dennis G. Zill
Loyola Marymount University
Michael R. Cullen
Late of Loyola Marymount University
By
Warren S. Wright
Loyola Marymount University
Carol D. Wright
* ; BROOKS/COLE
C EN G A G E Learning-
Australia • Brazil - Japan - Korea • Mexico • Singapore • Spain • United Kingdom • United States
Table of Contents
1 Introduction to Differential Equations 1
2 First-Order Differential Equations 27
3 Modeling with First-Order Differential Equations 86
4 Higher-Order Differential Equations 137
5 Modeling with Higher-Order Differential Equations 231
6 Series Solutions of Linear Equations 274
7 The Laplace Transform 352
8 Systems of Linear First-Order Differential Equations 419
9 Numerical Solutions of Ordinary Differential Equations 478
10 Plane Autonomous Systems 506
11 Fourier Series 538
12 Boundary-Value Problems in Rectangular Coordinates 586
13 Boundary-Value Problems in Other Coordinate Systems 675
14 Integral Transforms 717
15 Numerical Solutions of Partial Differential Equations 761
Appendix I Gamma function 783
Appendix II Matrices 785
1 Introduction to Differential Equations
1. Second order; linear
2. Third order; nonlinear because of (dy/dx)4
3. Fourth order; linear
4. Second order; nonlinear bccausc of cos(r + u)
5. Second order; nonlinear because of (dy/dx)2 or 1 + (dy/dx)2
6. Second order: nonlinear bccausc of R~
7. Third order: linear
8. Second order; nonlinear because of x2
9. Writing the differential equation in the form x(dy/dx) -f y2 = 1. we sec that it is nonlinear in y
because of y2. However, writing it in the form (y2 — 1 )(dx/dy) + x = 0, we see that it is linear in x.
10. Writing the differential equation in the form u(dv/du) + (1 + u)v = ueu wc see that it is linear in
v. However, writing it in the form (v + uv — ueu)(du/dv) + u — 0, we see that it, is nonlinear in ■Jill.
From y = e-*/2 we obtain y' = — \e~x'2. Then 2y' + y = —e~X//2 + e-x/2 = 0.
12. From y = | — |e-20* we obtain dy/dt = 24e-20t, so that
% + 20y = 24e~m + 20 - |e_20t) = 24.
clt \ 'o 5 /
13. R'om y = eix cos 2x we obtain y1 = 3e^x cos 2x — 2e3* sin 2a? and y” = 5e3,x cos 2x — 12e3,x sin 2x, so
that y" — (k/ + l?>y = 0.
14. From y = — cos:r ln(sec;r + tanrc) we obtain y’ — — 1 + sin.Tln(secx + tana:) and
y" = tan x + cos x ln(sec x + tan a?). Then y" -f y = tan x.
15. The domain of the function, found by solving x + 2 > 0, is [—2, oo). From y’ = 1 + 2(x + 2)_1/2 we
1
Exercises 1.1 Definitions and Terminology
have
{y - x)y' = (y - ®)[i + (20 + 2)_1/2]
= y — x + 2(y - x)(x + 2)-1/2
= y - x + 2[x + 4(z + 2)1/2 - a;] (a: + 2)_1/2
= y — x + 8(ac + 2)1;/i(rr + 2)~1/2 = y — x + 8.
An interval of definition for the solution of the differential equation is (—2, oo) because y
defined at x = —2.
16. Since tan:r is not defined for x = 7r/2 + mr, n an integer, the domain of y = 5t£v.
An interval of definition for the solution of the differential equation is (—7r/10,7T/10 . A:,
interval is (7r/10, 37t/10). and so on.
17. The domain of the function is {x \ 4 — x2 ^ 0} or {x\x ^ —2 or x ^ 2}. Prom y' — 2.:: -= -
we have
An interval of definition for the solution of the differential equation is (—2, 2). Other
(—oc,—2) and (2, oo).
18. The function is y — l/y/l — s in s. whose domain is obtained from 1 — sinx ^ 0 or . = 1 T
An interval of definition for the solution of the differential equation is (tt/2. 5tt/2 A:.. .
is (57r/2, 97r/2) and so on.
19. Writing ln(2X — 1) — ln(X — 1) = t and differentiating implicitly we obtain
2 dX 1 dX
2X - 1 dt X - l dt
{a; | 5x ^ tt/2 + 7i-7r} or {;r | x ^ tt/IO + mr/5}. From y' — 25sec2 §x we have
y' = 25(1 + tan2 5x) = 25 + 25 tan2 5a: = 25 + y2.
the domain is {z | x ^ tt/2 + 2?i7r}. From y' = —1(1 — sin x) 2 (— cos.x) we have
2y' = (1 — sin;r)_ ‘?/’2 cos# = [(1 — sin:r)~1//2]3cos:r - f/3cosx.
(2X - 1)(X - 1) dt
IX
— = -C2X - 1)(X - 1) = (X - 1)(1 - 2X .
2X - 2 - 2X + 1 dX _
2
Exercises 1.1 Definitions and Terminology
Exponentiating both sides of the implicit solution we obtain
2 X - 1
x
----- = el
X - l
2X - 1 = Xel - ef
(e* - 1) = (e‘ - 2)X
ef' — 1
- 4 -2
X =
e* - 2 '
-2
-4
Solving e* — 2 = 0 we get t = In2. Thus, the solution is defined on (—oc.ln2) or on (In2, oo).
The graph of the solution defined on (—oo,ln2) is dashed, and the graph of the solution defined on
(In 2. oc) is solid.
20. Implicitly differentiating the solution, we obtain y
—x2 dy — 2xy dx + y dy — 0
2 xy dx + (a;2 — y)dy = 0.
Using the quadratic formula to solve y2 — 2x2y — 1 = 0 for y, we get
y = (2x2 ± V4;c4 +4)/2 = a’2 ± vV 1 + 1. Thus, two explicit
solutions are y\ = x2 + \A'4 + 1 and y-2 = x2 — V.x4 + 1. Both
solutions are defined on (—oo. oc). The graph of yj (x) is solid and
the graph of y-2 is daalied.
21. Differentiating P = c\?} } ( l + cie^ we obtain
dP _ ( l + cie*) cie* - cie* • cie* _ Cie« [(l + cie‘) - cie4]
eft (1 + cie*)"
Ci
1 + CI&-
CiC
1 + ci ef
1 + cief
= P( 1 - P).
1 + cie(
2 PX ,2 ,2
22. Differentiating y = e~x / e: dt + c\ e~x we obtain
Jo
* f X *2 -r.2
/ e dt — 2c\xe = 1
Jo
y / = e -*2e r2
___ 2 r x J 2
2xe 2xe
2 rx +2
x e dt — 2cixe
Jo
—X
Substituting into the differential equation, we have
y' + 2xy = l — 2xe x I e* dt — 2cixe x +2xe x [ e* dt 4- 2cio;e x = 1.
Jo Jo
3